hdu-2602 背包问题 骨头收集者

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 89483 Accepted Submission(s): 36838

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

Sample Input

 
 
1
 5 10
 1 2 3 4 5
 5 4 3 2 1
 
 
Sample Output
 
 
14
 
 
#include 
using namespace std;
int max(int a, int b) {

    return a > b ? a : b;
}



int dp[1005][1005] ;
void main() {
    int t;
    int n;
    int v;
    cin >> t;
    int va[1000], vo[1000];
    while (t--) {
        cin >> n>>v;
            for (int i = 1; i <= n; i++)
                cin >> va[i];
            
            for (int i = 1; i <= n; i++) 
                cin >> vo[i];
            

            memset(dp, 0, sizeof(dp));


            for (int i = 1; i <= n; i++) {
                for (int j = 0; j <= v; j++)//要记得有容量为0的情况
                {
                    if (vo[i] > j) { dp[i][j] = dp[i - 1][j]; }
                    else if(vo[i] <= j) { dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - vo[i]] + va[i]); }


                }
            }
            cout << dp[n][v] << endl;
        }
    }